Class 11 Maths · Chapter 3
Short answer:
Class 11 Maths Chapter 3 Trigonometric Functions degree-radian conversion se shuru hokar general solutions of trigonometric equations tak jaata hai — radian measure, arc length (l = rθ), sabhi quadrants me trig functions ke signs, sum-difference (compound angle) formulas, aur sinθ = sinα / cosθ = cosα / tanθ = tanα ke general solutions, sab is chapter me aate hain. Neeche exercises 3.1 se 3.4 aur Miscellaneous Exercise ke representative solved questions step-by-step diye hain, taaki class 11 maths ncert solutions aur specifically class 11 maths chapter 3 trigonometric functions ncert solutions ek hi jagah, poore working ke saath mil jaayein.
Chapter 1 (Sets) aur Chapter 2 (Relations & Functions) ke baad, Chapter 3 Trigonometric Functions Class 11 Maths ka pehla formula-heavy chapter hai — aur isi wajah se ye poore Trigonometry (Class 11, Class 12, aur competitive exams) ki neev hai. Class 10 me tumne sirf right-angle triangle ke ratios (sin, cos, tan) padhe the; yahaan hum unhe circular functions ki tarah dekhte hain — jo kisi bhi real number x ke liye defined hain, sirf 0° se 90° ke beech nahi. Radian measure, unit circle par quadrant-wise signs (ASTC rule), compound angle formulas — sin(x±y), cos(x±y) — aur trigonometric equations ke general solutions: ye chaaron building blocks hain jo aage Class 11 ke baaki chapters aur Class 12 Calculus me baar-baar use hote hain. Ye chapter numerical-practice-heavy hai, isliye ratt-ne ke bajaye har identity ko derive karke samajhna zyada kaam aata hai.
Chapter 3 Summary — 5 Minute Revision
Is chapter me humne seekha: (1) degree aur radian ke beech conversion (radian = degree × π/180) aur arc-length formula l = rθ; (2) trigonometric functions ka domain saare real numbers tak extend karna aur unit circle par har quadrant me unke signs (ASTC — All Sine Tan Cos positive rule) pehchaanna; (3) sin²x + cos²x = 1 jaise Pythagorean identities aur negative-angle rules (sin, tan, cosec, cot odd hain; cos, sec even hain); (4) sum aur difference formulas — sin(x±y), cos(x±y), tan(x±y) — jinse double angle (sin2x, cos2x) aur triple angle (sin3x, cos3x) formulas derive hote hain, saath hi sum-to-product aur product-to-sum identities; aur (5) trigonometric equations (jaise sinx = sinα) ke general solution nikalna, jisme ek hi formula me har possible value of x cover ho jaati hai (nπ, 2nπ ± α, waghera, n ∈ Z). Jitna zyada Exercise 3.1 se 3.4 aur Miscellaneous ke sawaal practice karoge, identities prove karna aur equations solve karna utna hi natural lagega.
In-Text Questions — Solutions
Convert 40°20' into radian measure.
40°20' ko decimal degree me likhte hain: 40°20' = 40 + 20/60 degree = 40 + 1/3 = 121/3 degree.
radian = degree × π/180 = (121/3) × (π/180) = 121π/540 radian
Answer: 121π/540 radian.
Find the degree measure corresponding to 11π/16 radian.
degree = radian × 180/π = (11π/16) × (180/π) = (11×180)/16 = 1980/16 = 123.75°
0.75° = 0.75×60' = 45'. Answer: 123°45'.
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
In 60 seconds, wheel makes 360 revolutions ⇒ in 1 second, revolutions = 360/60 = 6.
1 complete revolution = 2π radian.
Angle turned in 1 second = 6 × 2π = 12π radian
Find the length of an arc of a circle of radius 5 cm subtending a central angle of 15°.
Pehle 15° ko radian me convert karo: 15 × π/180 = π/12 radian.
l = rθ = 5 × π/12 = 5π/12 cm
Answer: 5π/12 cm (≈ 1.31 cm).
If sinx = 3/5 and x lies in the second quadrant, find the values of the other five trigonometric functions.
sin²x + cos²x = 1 ⇒ cos²x = 1 - 9/25 = 16/25.
Q2 me cosine negative hota hai, isliye cosx = -4/5.
tanx = sinx/cosx = (3/5)/(-4/5) = -3/4
cosecx = 5/3, secx = -5/4, cotx = -4/3.
Find the value of sin 765°.
765° = 720° + 45° = 2×360° + 45°.
Trig functions ka period 360° hota hai, isliye poore revolutions hata do:
sin 765° = sin 45° = 1/√2
Find the value of cos(-1710°).
cos even function hai: cos(-1710°) = cos(1710°).
1710° = 4×360° + 270° = 1440° + 270°.
cos(1710°) = cos 270° = 0
Answer: 0.
Prove that cos(π/4 + x) + cos(π/4 - x) = √2 cosx.
Sum-to-product identity: cosA + cosB = 2 cos((A+B)/2) cos((A-B)/2). Yahaan A = π/4+x, B = π/4-x.
(A+B)/2 = π/4, (A-B)/2 = x
LHS = 2 cos(π/4) cos(x) = 2×(1/√2)×cosx = √2 cosx = RHS
Hence proved.
Find the value of sin 75°.
75° = 45° + 30°. Sum formula use karo:
sin75° = sin45°cos30° + cos45°sin30° = (1/√2)(√3/2) + (1/√2)(1/2)
= (√3+1)/(2√2) = (√6+√2)/4
Find the value of tan(13π/12).
13π/12 = π + π/12. Aur tan(π + θ) = tanθ (period π).
tan(13π/12) = tan(π/12) = tan15° = 2 - √3
(tan15° ko tan(45°-30°) se derive kiya jaata hai: (1-1/√3)/(1+1/√3) = 2-√3.)
Prove that sin3x = 3sinx - 4sin³x.
sin3x = sin(2x+x) = sin2x·cosx + cos2x·sinx
= (2sinxcosx)·cosx + (1-2sin²x)·sinx
= 2sinxcos²x + sinx - 2sin³x = 2sinx(1-sin²x) + sinx - 2sin³x
= 2sinx - 2sin³x + sinx - 2sin³x = 3sinx - 4sin³x
Hence proved.
Find the general solution of sinx = -√3/2.
sin(-π/3) = -√3/2, isliye α = -π/3 lete hain.
sinx = sinα ⇒ x = nπ + (-1)ⁿα = nπ + (-1)ⁿ(-π/3), n ∈ Z
Answer: x = nπ - (-1)ⁿ π/3, n ∈ Z.
Find the general solution of tan2x = -1.
tan(-π/4) = -1, isliye tan2x = tan(-π/4).
2x = nπ + (-π/4) ⇒ x = nπ/2 - π/8, n ∈ Z
Write the principal solutions of sinx = -1/2.
Sine negative hoti hai third aur fourth quadrant me. Reference angle = π/6.
Q3: x = π + π/6 = 7π/6 | Q4: x = 2π - π/6 = 11π/6
Principal solutions: x = 7π/6, 11π/6.
Prove that sin(n+1)x·sin(n+2)x + cos(n+1)x·cos(n+2)x = cosx.
cos(A-B) = cosA cosB + sinA sinB formula ko reverse me pehchaano. Yahaan A = (n+2)x, B = (n+1)x.
cosA cosB + sinA sinB = cos(A-B) = cos((n+2)x - (n+1)x) = cosx
Hence proved.
Poore Class 11 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q30)
(Ex 3.1) Find the radian measure corresponding to 240°.
radian = 240 × π/180 = 4π/3
(Ex 3.1) Find the radian measure corresponding to -47°30'.
-47°30' = -(47 + 30/60)° = -(95/2)°.
radian = -(95/2) × π/180 = -95π/360 = -19π/72
(Ex 3.1) Find the degree measure corresponding to 7π/6 radian.
degree = (7π/6) × (180/π) = 7×30 = 210°
(Ex 3.1) Find the degree measure corresponding to -4 radian (use π = 22/7).
degree = -4 × 180/π = -4 × (180×7/22) = -5040/22 ≈ -229.09°
0.09° ≈ 5'27". Answer ≈ -229°5'27" (π = 22/7 use karne se approximate).
(Ex 3.1) A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
1 second me revolutions = 360/60 = 6.
Angle = 6 × 2π = 12π radian
(Ex 3.1) The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (use π = 3.14)
60 minute me minute hand poora circle (2π radian) cover karta hai. 40 minute me:
θ = (40/60) × 2π = 4π/3 radian
l = rθ = 1.5 × 4π/3 = 2π = 2×3.14 = 6.28 cm
(Ex 3.1) Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (use π = 22/7).
θ = l/r = 22/100 = 0.22 radian
degree = 0.22 × 180/π = 0.22 × 1260/22 = 12.6° = 12°36'
(Ex 3.2) If cosx = -1/2 and x lies in the third quadrant, find the values of the other five trigonometric functions.
sin²x = 1 - cos²x = 1 - 1/4 = 3/4. Q3 me sine negative: sinx = -√3/2.
tanx = sinx/cosx = (-√3/2)/(-1/2) = √3
cosecx = -2/√3 = -2√3/3, secx = -2, cotx = 1/√3 = √3/3.
(Ex 3.2) If cotx = 3/4 and x lies in the third quadrant, find the values of the other five trigonometric functions.
1 + cot²x = cosec²x ⇒ cosec²x = 1 + 9/16 = 25/16. Q3 me sine negative: cosecx = -5/4.
sinx = 1/cosecx = -4/5, cosx = cotx × sinx = (3/4)(-4/5) = -3/5
tanx = 1/cotx = 4/3, secx = 1/cosx = -5/3.
Check: sin²x + cos²x = 16/25 + 9/25 = 1 ✓
(Ex 3.2) Find the value of sin(-11π/3).
-11π/3 + 2×(2π) = -11π/3 + 12π/3 = π/3 (period 2π use karke).
sin(-11π/3) = sin(π/3) = √3/2
(Ex 3.2) Find the value of cot(-15π/4).
cot ka period π hai. -15π/4 + 4π = -15π/4 + 16π/4 = π/4.
cot(-15π/4) = cot(π/4) = 1
(Ex 3.2) Find the value of tan(19π/3).
tan ka period π hai. 19π/3 - 6π = 19π/3 - 18π/3 = π/3.
tan(19π/3) = tan(π/3) = √3
(Ex 3.3) Prove that sin²(π/6) + cos²(π/3) - tan²(π/4) = -1/2.
sin(π/6) = 1/2 ⇒ sin² = 1/4. cos(π/3) = 1/2 ⇒ cos² = 1/4. tan(π/4) = 1 ⇒ tan² = 1.
LHS = 1/4 + 1/4 - 1 = 1/2 - 1 = -1/2 = RHS
Hence proved.
(Ex 3.3) Prove that 2sin²(π/6) + cosec²(7π/6)cos²(π/3) = 3/2.
sin(π/6) = 1/2 ⇒ 2×1/4 = 1/2.
sin(7π/6) = sin(π+π/6) = -sin(π/6) = -1/2 ⇒ cosec(7π/6) = -2 ⇒ square = 4.
cos(π/3) = 1/2 ⇒ square = 1/4.
LHS = 1/2 + 4×1/4 = 1/2 + 1 = 3/2 = RHS
(Ex 3.3) Prove that sin(x+y)/sin(x-y) = (tanx+tany)/(tanx-tany).
sin(x+y) = sinxcosy + cosxsiny; sin(x-y) = sinxcosy - cosxsiny. Dono ko cosx·cosy se divide karo:
sin(x+y)/(cosxcosy) = tanx + tany, sin(x-y)/(cosxcosy) = tanx - tany
⇒ sin(x+y)/sin(x-y) = (tanx+tany)/(tanx-tany)
Hence proved.
(Ex 3.3) Prove that tan(π/4 + x)/tan(π/4 - x) = ((1+tanx)/(1-tanx))².
tan(π/4+x) = (1+tanx)/(1-tanx) aur tan(π/4-x) = (1-tanx)/(1+tanx) (standard sum/difference formula).
Ratio = [(1+tanx)/(1-tanx)] ÷ [(1-tanx)/(1+tanx)] = [(1+tanx)/(1-tanx)]²
Hence proved.
(Ex 3.3) Prove that (sin5x - 2sin3x + sinx)/(cos5x - cosx) = tanx.
Numerator: sin5x + sinx = 2sin3x cos2x (sum-to-product), isliye sin5x-2sin3x+sinx = 2sin3xcos2x - 2sin3x = 2sin3x(cos2x-1) = 2sin3x(-2sin²x) = -4sin3x·sin²x.
Denominator: cos5x - cosx = -2sin3x·sin2x.
Ratio = (-4sin3x·sin²x)/(-2sin3x·sin2x) = 2sin²x/sin2x = 2sin²x/(2sinxcosx) = sinx/cosx = tanx
Hence proved.
(Ex 3.3) Prove that cos4x = 1 - 8sin²x cos²x.
cos4x = 1 - 2sin²(2x) [cos2θ = 1-2sin²θ, θ=2x]
= 1 - 2(2sinxcosx)² = 1 - 2×4sin²xcos²x = 1 - 8sin²xcos²x
Hence proved.
(Ex 3.3) Prove that cos6x = 32cos⁶x - 48cos⁴x + 18cos²x - 1.
cos6x = cos(3·2x) = 4cos³(2x) - 3cos(2x). Let c = cos2x = 2cos²x - 1, u = cos²x, so c = 2u-1.
4c³ - 3c = 4(2u-1)³ - 3(2u-1)
(2u-1)³ = 8u³-12u²+6u-1, isliye 4(2u-1)³ = 32u³-48u²+24u-4.
-3(2u-1) = -6u+3 ⇒ Total = 32u³-48u²+18u-1 = 32cos⁶x-48cos⁴x+18cos²x-1
Hence proved.
(Ex 3.3) Prove that sinx + sin3x + sin5x + sin7x = 4cosx cos2x sin4x.
Group karo: (sinx+sin7x) + (sin3x+sin5x)
= 2sin4xcos3x + 2sin4xcosx = 2sin4x(cos3x+cosx) = 2sin4x×2cos2xcosx
= 4sin4x·cos2x·cosx = 4cosx cos2x sin4x
Hence proved.
(Ex 3.4) Find the principal solutions of the equation tanx = -1/√3.
Tan negative hota hai Q2 aur Q4 me. Reference angle π/6 (tan π/6 = 1/√3).
Q2: x = π - π/6 = 5π/6 | Q4: x = 2π - π/6 = 11π/6
Principal solutions: x = 5π/6, 11π/6.
(Ex 3.4) Find the general solution of the equation sinx = 0.
sinx = 0 ⇒ x = nπ, n ∈ Z
(Ex 3.4) Find the general solution of the equation cosx = 0.
cosx = 0 ⇒ x = (2n+1)π/2, n ∈ Z
(Ex 3.4) Find the general solution of the equation cosecx = -2.
cosecx = -2 ⇒ sinx = -1/2 = sin(-π/6).
x = nπ + (-1)ⁿ(-π/6) = nπ - (-1)ⁿπ/6, n ∈ Z
(Ex 3.4) Find the general solution of the equation sin2x - sinx = 0.
sin2x = 2sinxcosx, isliye 2sinxcosx - sinx = 0 ⇒ sinx(2cosx-1) = 0.
sinx = 0 ⇒ x = nπ | cosx = 1/2 ⇒ x = 2nπ ± π/3, n ∈ Z
(Ex 3.4) Find the general solution of the equation cos4x = cosx.
cos4x - cosx = 0 ⇒ -2sin(5x/2)sin(3x/2) = 0 (sum-to-product, A=4x, B=x).
sin(5x/2) = 0 ⇒ x = 2nπ/5 | sin(3x/2) = 0 ⇒ x = 2mπ/3, n,m ∈ Z
(Ex 3.4) Find the general solution of the equation sin2x = cos3x.
cos3x = sin(π/2-3x) ke roop me likho, isliye sin2x = sin(π/2-3x), yaani cos(π/2-2x) = cos3x form use karke:
π/2 - 2x = 2nπ ± 3x
Case (+): π/2-2x = 2nπ+3x ⇒ x = π/10 - 2nπ/5.
Case (-): π/2-2x = 2nπ-3x ⇒ x = 2nπ - π/2, n ∈ Z.
(Misc) Prove that (cosx+cosy)² + (sinx-siny)² = 4cos²((x+y)/2).
Expand: cos²x+2cosxcosy+cos²y+sin²x-2sinxsiny+sin²y
= (cos²x+sin²x) + (cos²y+sin²y) + 2(cosxcosy-sinxsiny) = 1+1+2cos(x+y)
= 2(1+cos(x+y)) = 2×2cos²((x+y)/2) = 4cos²((x+y)/2)
Hence proved.
(Misc) If tanx = -4/3 and x lies in the second quadrant, find sin(x/2), cos(x/2), tan(x/2).
Q2 me sinx = 4/5, cosx = -3/5 (ratio -4/3 confirm hoti hai). x/2 lies in Q1 (x ∈ (π/2,π) ⇒ x/2 ∈ (π/4,π/2)), sab positive.
sin²(x/2) = (1-cosx)/2 = (1+3/5)/2 = 4/5 ⇒ sin(x/2) = 2/√5
cos²(x/2) = (1+cosx)/2 = (1-3/5)/2 = 1/5 ⇒ cos(x/2) = 1/√5
tan(x/2) = sin(x/2)/cos(x/2) = 2
(Misc) Prove that cos²2x - cos²6x = sin4x sin8x.
Identity: cos²A - cos²B = sin(A+B)sin(B-A). Yahaan A = 2x, B = 6x.
cos²2x - cos²6x = sin(2x+6x)sin(6x-2x) = sin8x sin4x
Hence proved.
Important Equations — Ek Nazar Me
| Category | Formula |
|---|---|
| Degree ↔ Radian | radian = degree × π/180 | degree = radian × 180/π |
| Arc length | l = rθ (θ radian me, r aur l same unit me) |
| Pythagorean identities | sin²x + cos²x = 1 | 1 + tan²x = sec²x | 1 + cot²x = cosec²x |
| Negative angle | sin(-x) = -sinx, tan(-x) = -tanx, cosec(-x) = -cosecx, cot(-x) = -cotx (odd) | cos(-x) = cosx, sec(-x) = secx (even) |
| Sum formulas | sin(x+y) = sinx cosy + cosx siny | cos(x+y) = cosx cosy - sinx siny | tan(x+y) = (tanx + tany)/(1 - tanx tany) |
| Difference formulas | sin(x-y) = sinx cosy - cosx siny | cos(x-y) = cosx cosy + sinx siny | tan(x-y) = (tanx - tany)/(1 + tanx tany) |
| Double angle | sin2x = 2sinx cosx | cos2x = cos²x - sin²x = 2cos²x - 1 = 1 - 2sin²x | tan2x = 2tanx/(1-tan²x) |
| Triple angle | sin3x = 3sinx - 4sin³x | cos3x = 4cos³x - 3cosx | tan3x = (3tanx - tan³x)/(1-3tan²x) |
| Sum-to-product | sinA + sinB = 2 sin((A+B)/2) cos((A-B)/2) | sinA - sinB = 2 cos((A+B)/2) sin((A-B)/2) | cosA + cosB = 2 cos((A+B)/2) cos((A-B)/2) | cosA - cosB = -2 sin((A+B)/2) sin((A-B)/2) |
| Product-to-sum | 2 sinx cosy = sin(x+y) + sin(x-y) | 2 cosx siny = sin(x+y) - sin(x-y) | 2 cosx cosy = cos(x+y) + cos(x-y) | 2 sinx siny = cos(x-y) - cos(x+y) |
| General solutions | sinθ = sinα ⇒ θ = nπ + (-1)ⁿα | cosθ = cosα ⇒ θ = 2nπ ± α | tanθ = tanα ⇒ θ = nπ + α (n ∈ Z) |
| Half-angle (cos in terms of half) | sin²(x/2) = (1-cosx)/2 | cos²(x/2) = (1+cosx)/2 | tan(x/2) = sinx/(1+cosx) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Degree se radian aur radian se degree convert karte waqt π/180 aur 180/π ko ulta laga dena — hamesha check karo target unit kya chahiye.
- Quadrant-wise signs (ASTC rule) galat yaad rakhna, especially Q2 aur Q4 me tan/cot ka sign confuse kar dena.
- 'Principal solution' aur 'general solution' ko mix kar dena — general solution me hamesha n ∈ Z wala term hona chahiye, sirf ek value nahi.
- sin(x+y) ko galti se sinx + siny samajh lena — trigonometric functions linear nahi hote, isliye distribute nahi hota.
- cos2x ke teen alag forms (1-2sin²x, 2cos²x-1, cos²x-sin²x) me se galat form choose karna jab question specifically sinx ya cosx me answer maange.
- Negative angle rules bhool jaana — sin, tan, cosec, cot odd functions hain (sin(-x) = -sinx) jabki cos, sec even hain (cos(-x) = cosx); sign galat lagane se poora answer ulta ho jaata hai.
Board-Style Important Questions
- Kisi diye gaye degree measure ko radian me (ya radian ko degree me) convert karna — seedha formula radian = degree × π/180 ka direct application, is chapter ka sabse basic aur common question pattern.
- Ek trigonometric ratio (jaise sinx ya cotx) aur quadrant diya ho, baaki paanch trigonometric functions ki values nikalna — Pythagorean identity aur quadrant-sign dono test hote hain.
- Arc length ya radius se related short problem — l = rθ formula use karke angle ya arc length nikalna.
- Sum/difference ya double-angle formulas use karke ek trigonometric identity prove karna (jaise sin3x = 3sinx - 4sin³x type) — is chapter ka signature identity-proof pattern.
- Diye gaye trigonometric equation (jaise sin2x = cos3x ya tan2x = -1) ka general solution nikalna, poora working dikhate hue.
- Multi-step identity (jaise cos6x ka expansion, ya sum-to-product wali long-form identity) prove karna, jisme intermediate double/triple angle formulas chahiye — lambi, zyada steps wale proof questions.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths Chapter 3 Trigonometric Functions NCERT Solutions kahan se step-by-step milengi?
Upar diye gaye intext aur exercise sections me Ex 3.1 se 3.4 aur Miscellaneous Exercise ke representative sawaal full step-by-step working ke saath diye hain — radian conversion se lekar general solutions tak. Ye class 11 maths chapter 3 trigonometric functions ncert solutions ka core coverage hai.
Kya Class 11 Maths Chapter 1 Sets is chapter se directly related hai?
Chapter 1 (Sets) trigonometric equations ke solution-sets likhne me indirectly kaam aata hai — jaise general solution ko set-builder form me express karna. Waise Class 11 Maths NCERT Solutions series me har chapter connected building block hai, par Sets ka direct formula-overlap Trigonometric Functions se nahi hai.
Is chapter ke saare formulas ek jagah — class 11 maths formulas pdf download kaise karein?
Upar 'formulas' table me is chapter ke saare key formulas (degree-radian, Pythagorean identities, sum-difference, double/triple angle, general solutions) diye hain. In sabko compile karke apna khud ka formula-sheet PDF bana sakte ho revision ke liye.
Class 11 Maths syllabus me Trigonometric Functions ka weightage kya hai?
Exact marks-weightage numbers is spec me nahi diye gaye hain kyunki sources ke beech figures conflict karte hain aur weightage session-to-session change hoti rehti hai — accurate weightage janne ke liye CBSE ke official curriculum document ko directly check karo.
Class 11 Maths NCERT book PDF download kahan se karein?
Official NCERT textbook PDF ncert.nic.in par free available hota hai — 'class 11 maths ncert book pdf download' search karke NCERT ki official site se hi download karna best rehta hai, taaki content authentic ho.
Kya is chapter ke liye class 11 maths exemplar solutions bhi zaroori hain?
Haan — NCERT textbook exercises basics clear karte hain, par Exemplar me thode tougher aur multi-step identity/general-solution questions hote hain jo board-level ya competitive prep ke liye extra practice dete hain. Trigonometric Functions jaisa formula-heavy chapter Exemplar practice se aur mazboot hota hai.
Class 11 Maths — Saare Chapters
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